If a, b and 8 are a GP and a, 8 and b are an AP, find a and b.

For three terms a, b and c in a GP, ac = b^2. And for
three terms a, b and c which are in an AP a+c= 2b.

As a, b
and 8 are in a GP 8a = b^2

And as a, 8 and b are in an AP
a+b = 2*8= 16

Now we will use the two
equations:

8a = b^2 => a = b^2 /
8

substitute this in a+ b = 16 => b^2/ 8 + b
=16

=> b^2 + 8 b =
8*16

=> b^2 + 8b – 128
=0

=> b^2 + 16b- 8b -128
=0

=> b(b+16) –
8(b+16)=0

=> (b-8)(b+16)
=0

=> b = 8 or b=
-16

Now a+b=16

=> a +8
= 16

=> a = 8

This
gives an AP with difference 0 and common ratio 1.

For a-16
=16

=> a = 32

Here the
common difference is -24 and common ratio is
-2

The required sets of a and b are (32, -16)
and (8,8)